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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><span class="process-math">\(r_1-r_2=1\)</span> corresponds to Case 3 of the theorem. According to the theorem, the first solution is given by</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_20.html ./knowl/eq5_21.html">
\begin{equation}
y_1=x^{r_1} \sum_{n=0}^{\infty} a_n x^n=\sum_{n=0}^{\infty} a_n x^{n+1/2} \quad \textrm{with}~a_0=1.\tag{5.5.11}
\end{equation}
</div>
<p class="continuation">Substituting (<a href="" class="xref" data-knowl="./knowl/eq5_20.html" title="Equation 5.5.11">(5.5.11)</a>) into (<a href="" class="xref" data-knowl="./knowl/eq5_21.html" title="Equation 5.5.10">(5.5.10)</a>):</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_20.html ./knowl/eq5_21.html">
\begin{equation*}
\begin{aligned}
&amp;x^2 \sum_{n=0}^{\infty} (n+1/2) (n-1/2) a_n x^{n-3/2}+x \sum_{n=0}^{\infty}(n+1/2) a_n x^{n-1/2}+(x^2-1/4) \sum_{n=0}^{\infty} a_n x^{n+1/2}=0,\\
&amp;\to \sum_{n=0}^{\infty} (n^2-1/4) a_n x^{n+1/2}+\sum_{n=0}^{\infty} (n+1/2) a_n x^{n+1/2}+\sum_{n=0}^{\infty} a_n x^{n+5/2}-\frac{1}{4} \sum_{n=0}^{\infty} a_n x^{n+1/2}=0.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Dividing both sides by the factor <span class="process-math">\(x^{1/2}\text{:}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_20.html ./knowl/eq5_21.html">
\begin{equation*}
\begin{aligned}
&amp;\to \sum_{n=0}^{\infty} (n^2-1/4) a_n x^n+\sum_{n=0}^{\infty} (n+1/2) a_n x^n+\sum_{n=0}^{\infty} a_n x^{n+2}-\frac{1}{4} \sum_{n=0}^{\infty} a_n x^n=0\\
&amp;\to \sum_{n=0}^{\infty} [n^2-\frac{1}{4}+n+\frac{1}{2}-\frac{1}{4}] a_n x^n+\sum_{n=0}^{\infty} a_n x^{n+2}=0\\
&amp;\to \sum_{n=0}^{\infty} n(n+1) a_n x^n+\sum_{k=2}^{\infty}a_{k-2} x^k=0\\
&amp;\to 0+1 \times 2 \times a_1 x+\sum_{n=2}^{\infty} n(n+1) a_n x^n+\sum_{n=2}^{\infty} a_{n-2} x^n=0\\
&amp;\to 2 a_1 x+\sum_{n=2}^{\infty} [n(n+1) a_n+a_{n-2}] x^n=0\\
&amp;\to a_1=0,\quad n(n+1)a_n+a_{n-2}=0,\\
&amp;\to  a_1=0, \quad a_n=-\frac{a_{n-2}}{n(n+1)},\quad n \geq 2.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Based on this recurrence relation, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_20.html ./knowl/eq5_21.html">
\begin{equation*}
\begin{aligned}
&amp;n=2:\quad a_2=-\frac{a_0}{3 \cdot 2}=-\frac{1}{3!},\\
&amp;n=3: \quad a_3=0,\\
&amp;n=4: \quad a_4=-\frac{a_2}{5\cdot 4}=+\frac{1}{5!},\\
&amp;n=5:\quad a_5=0,\\
&amp;n=6: \quad a_6=-\frac{a_4}{7 \cdot 6}=-\frac{1}{7!},\\
&amp;\cdots\\
&amp;a_{2k+1}=0,\\
&amp;a_{2k}=(-1)^k \frac{1}{(2k+1)!}.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_20.html ./knowl/eq5_21.html">
\begin{equation*}
\begin{aligned}
y_1&amp;=\sum_{n=0}^{\infty} a_n x^{n+1/2}=\sum_{k=0}^{\infty} a_{2k} x^{2k+1/2}
+\sum_{k=0}^{\infty} a_{2k+1}x^{2k+1+1/2}\\
&amp;=\sum_{n=0}^{\infty} (-1)^k \frac{x^{2k+1/2}}{(2k+1)!}=x^{-1/2} \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}\\
&amp;=x^{-1/2} \sin x.
\end{aligned}
\end{equation*}
</div>
<span class="incontext"><a href="sec5_5.html#p-236" class="internal">in-context</a></span>
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